Boxes can contain 0 balls in them too.
2007-10-11 10:31:24 · 16 answers · asked by dan_steinke 1 in Science & Mathematics ➔ Mathematics
135 different ways
2007-10-11 10:34:25 · answer #1 · answered by The Quikstr 3 · 0⤊ 1⤋
I will assume that the balls are indistinguishable, but that the boxes are distinguishable.
Imagine having the 15 balls set out in a row. You have two sticks, labeled A and B. You set down stick A first, either to the left of all the balls, in between two balls, or to the right of all the balls. All of the balls to the left of A will go in box #1. Now lay down stick B, either in the same place as stick A, or to its right. All of the balls between A and B go in box #2, and all of the balls to the right of B go in box #3.
How many ways can you lay down A and B? Well, if A is put to the left of all the balls, there are 16 places to put B. If A is put to the right of the first ball, there are 15 places to put B. If there are two balls to the left of A, then there are 14 places for B. And so on. So the total number of ways to place A and B is 15+14+13+...+3+2+1 = 120.
And since each placement of the two sticks uniquely specifies one dividing up of the balls, there are 120 ways to place the balls in the three boxes.
2007-10-11 11:42:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Assuming the first box contains 0 then the other 2 can have 15 possibilities 0,15 1,14 2,13 .... 13,2,14,1 15,0
Assuming the first box contains 1 then the other 2 can have 14 possibilities 0,14, 1,13 2,12 .... 12,2 13,1 14,0
-----etc
Assuming the first box contains 14 then the other 2 can have 2 possibilities 0,1 and 1,0
Assuming the first box contains 15 then the other 2 can have 1 possibility 0,0.
15+14+13+12+11+10+9+8+7+6+5+4+3+2+1= 120
2007-10-11 10:49:41 · answer #3 · answered by Barkley Hound 7 · 0⤊ 0⤋
I GOT THE ANWSER!!!------since the boxes can contain no balls, you put the 15 balls in one box, the "boxes" can remain empty and the box has all 15.
WOW THAT TOOK ME A WHILE
2007-10-11 10:40:50 · answer #4 · answered by Japhy 3 · 0⤊ 0⤋
For each ball, you have 3 options (you could put them in any box).
for 15 balls, there are 3 option for each ball, so 3 to the power of 15, or 14348907, is the total number of ways you could arrange the balls.
This is assuming that all of the balls are unique.
2007-10-11 10:38:09 · answer #5 · answered by Anonymous · 2⤊ 1⤋
2730. You need a TI30xa calculator. Hit 15, 2nd, 9, 3, enter. The way you get it is 15x14x13. Its similar to 15! (15 factorial) but in this case its not 15x14x13x12x11x10x9x8x7x6x5x4x3x2x1 but its 15x14x13
2007-10-11 10:42:46 · answer #6 · answered by cptnkeyes 3 · 0⤊ 0⤋
1st ball can be put into any 1 of 3 boxes in 3 ways.
2nd ball can be put into any 1 of 3 boxes in 3 ways.
Each way of putting 1st ball can be associated with each way of putting 2nd ball.
So, total no. of ways of putting 2 balls in 3 boxes = 3^2.
Counting this way, total no. of putting 15 balls in 3 boxes
= 3^(15)
= 14348907.
2007-10-11 10:40:35 · answer #7 · answered by Madhukar 7 · 1⤊ 1⤋
the right answer is 3375, which is 15X15X15. The reason it is NOT 15 factorial (limit 3) or 15X14X13 is because you can put all the balls in one box.
2007-10-11 10:36:46 · answer #8 · answered by Mike 7 · 0⤊ 1⤋
If you have 15 balls and that is your only concern..I say get out more often..
2007-10-11 10:36:31 · answer #9 · answered by dusty 2 · 0⤊ 1⤋
I'm guess 15 to the 3rd power. Just a guess.
2007-10-11 10:33:41 · answer #10 · answered by Anonymous · 0⤊ 1⤋